1. Integer Linear Programming
Chapter 6
Integer Linear Programming

2. Introduction ILPs occur frequently…
When one or more variables in an LP problem must assume an integer value we have an Integer Linear Programming (ILP) problem.
ILPs occur frequently…
Scheduling workers
Manufacturing airplanes
Integer variables also allow us to build more accurate models for a number of common business problems.

3. Integrality Conditions
MAX: 350X X2 } profit
S.T.: 1X1 + 1X2 <= 200 } pumps
9X1 + 6X2 <= 1566 } labor
12X1 + 16X2 <= 2880 } tubing
X1, X2>= 0 } nonnegativity
X1, X2 must be integers } integrality
Integrality conditions are easy to state but make the problem much more difficult (and sometimes impossible) to solve.

4. Relaxation Original ILP LP Relaxation MAX: 2X1 + 3X2
S.T.: X1 + 3X2 <= 8.25
2.5X1 + X2 <= 8.75
X1, X2 >= 0
X1, X2 must be integers
LP Relaxation

5. Integer Feasible vs. LP Feasible RegionX1 X2
Integer Feasible Solutions 1 2 3 4

6. Solving ILP Problems
When solving an LP relaxation, sometimes you “get lucky” and obtain an integer feasible solution.
This was the case in the original Blue Ridge Hot Tubs problem in earlier chapters.
But what if we reduce the amount of labor available to 1520 hours and the amount of tubing to 2650 feet?
See file Fig6-2.xls

7. Bounds
The optimal solution to an LP relaxation of an ILP problem gives us a bound on the optimal objective function value.
For maximization problems, the optimal relaxed objective function values is an upper bound on the optimal integer value.
For minimization problems, the optimal relaxed objective function values is a lower bound on the optimal integer value.

8. Rounding
It is tempting to simply round a fractional solution to the closest integer solution.
In general, this does not work reliably:
The rounded solution may be infeasible.
The rounded solution may be suboptimal.

9. How Rounding Down Can Result in an Infeasible SolutionX1 X2 1 2 3 4
optimal relaxed solution
infeasible solution obtained by rounding down

10. Branch-and-Bound
The Branch-and-Bound (B&B) algorithm can be used to solve ILP problems.
Requires the solution of a series of LP problems termed “candidate problems”.
Theoretically, this can solve any ILP.
Practically, it often takes LOTS of computational effort (and time).

11. Stopping Rules
Because B&B can take so long, most ILP packages allow you to specify a suboptimality tolerance factor.
This allows you to stop once an integer solution is found that is within some % of the global optimal solution.
Bounds obtained from LP relaxations are helpful here.
Example
LP relaxation has an optimal obj. value of $64,306.
95% of $64,306 is $61,090.
Thus, an integer solution with obj. value of $61,090 or better must be within 5% of the optimal solution.

12. Using Solver
Let’s see how to specify integrality conditions and suboptimality tolerances using Solver…
See file Fig6-8.xls

13. An Employee Scheduling Problem: Air-Express
Day of Week Workers Needed
Sunday 18
Monday 27
Tuesday 22
Wednesday 26
Thursday 25
Friday 21
Saturday 19
Shift Days Off Wage
1 Sun & Mon $680
2 Mon & Tue $705
3 Tue & Wed $705
4 Wed & Thr $705
5 Thr & Fri $705
6 Fri & Sat $680
7 Sat & Sun $655

14. Defining the Decision Variables
X1 = the number of workers assigned to shift 1
X2 = the number of workers assigned to shift 2
X3 = the number of workers assigned to shift 3
X4 = the number of workers assigned to shift 4
X5 = the number of workers assigned to shift 5
X6 = the number of workers assigned to shift 6
X7 = the number of workers assigned to shift 7

15. Defining the Objective Function
Minimize the total wage expense.
MIN: 680X1 +705X2 +705X3 +705X4 +705X5 +680X6 +655X7

16. Defining the Constraints
Workers required each day
0X1+ 1X2+ 1X3+ 1X4+ 1X5+ 1X6+ 0X7 >= 18 } Sunday
0X1+ 0X2+ 1X3+ 1X4+ 1X5+ 1X6+ 1X7 >= 27 } Monday
1X1+ 0X2+ 0X3+ 1X4+ 1X5+ 1X6+ 1X7 >= 22 }Tuesday
1X1+ 1X2+ 0X3+ 0X4+ 1X5+ 1X6+ 1X7 >= 26 } Wednesday
1X1+ 1X2+ 1X3+ 0X4+ 0X5+ 1X6+ 1X7 >= 25 } Thursday
1X1+ 1X2+ 1X3+ 1X4+ 0X5+ 0X6+ 1X7 >= 21 } Friday
1X1+ 1X2+ 1X3+ 1X4+ 1X5+ 0X6+ 0X7 >= 19 } Saturday
Nonnegativity & integrality conditions
Xi >= 0 and integer for all i

17. Implementing the Model
See file Fig6-14.xls

18. Binary Variables
Binary variables are integer variables that can assume only two values: 0 or 1.
These variables can be useful in a number of practical modeling situations….

19. A Capital Budgeting Problem: CRT Technologies
Expected NPV
Project (in $000s) Year 1 Year 2 Year 3 Year 4 Year 5
1 $141 $75 $25 $20 $15 $10
2 $187 $90 $35 $0 $0 $30
3 $121 $60 $15 $15 $15 $15
4 $83 $30 $20 $10 $5 $5
5 $265 $100 $25 $20 $20 $20
6 $127 $50 $20 $10 $30 $40
Capital (in $000s) Required in
The company has $250,000 available to invest in new projects. It has budgeted $75,000 for continued support for these projects in year 2 and $50,000 per year for years 3, 4, and 5.
Unused funds in any year cannot be carried over.

20. Defining the Decision Variables

21. Defining the Objective Function
Maximize the total NPV of selected projects.
MAX: 141X X X3
+ 83X X X6

22. Defining the Constraints
Capital Constraints
75X1 + 90X2 + 60X3 + 30X X5 + 50X6 <= } year 1
25X1 + 35X2 +15X3 + 20X X5 + 20X6 <= } year 2
20X1 + 0X X3 + 10X4 + 20X5 + 10X6 <= } year 3
15X1 + 0X X3 + 5X X5 + 30X6 <= } year 4
10X1 +30X2 +15X X X5 + 40X6 <= } year 5
Binary Constraints
All Xi must be binary

23. Implementing the Model
See file Fig6-17.xls

24. Binary Variables & Logical Conditions
Binary variables are also useful in modeling a number of logical conditions.
Of projects 1, 3 & 6, no more than one may be selected
X1 + X3 + X6 <= 1
Of projects 1, 3 & 6, exactly one must be selected
X1 + X3 + X6 = 1
Project 4 cannot be selected unless project 5 is also selected
X4 – X5 <= 0

25. The Fixed-Charge Problem
Many decisions result in a fixed or lump-sum cost being incurred:
The cost to lease, rent, or purchase a piece of equipment or a vehicle that will be required if a particular action is taken.
The setup cost required to prepare a machine or to produce a different type of product.
The cost to construct a new production line that will be required if a particular decision is made.
The cost of hiring additional personnel that will be required if a particular decision is made.

26. Example Fixed-Charge Problem: Remington Manufacturing
Hours Required By:
Operation Prod. 1 Prod. 2 Prod. 3 Hours Available
Machining
Grinding
Assembly
Unit Profit $48 $55 $50
Setup Cost $1000 $800 $900

27. Defining the Objective Function
Maximize total profit.
MAX: 48X1 + 55X2 + 50X3 – 1000Y1 – 800Y2 – 900Y3

28. Defining the Decision Variables
Xi = the amount of product i to be produced, i = 1, 2, 3

29. Defining the Constraints
Resource Constraints
2X1 + 3X2 + 6X3 <= 600 } machining
6X1 + 3X2 + 4X3 <= 300 } grinding
5X1 + 6X2 + 2X3 <= 400 } assembly
Binary Constraints
All Yi must be binary
Nonnegativity conditions
Xi >= 0, i = 1, 2, ..., 6
Is there a missing link?

30. Defining the Constraints (cont’d)
Linking Constraints (with “Big M”)
X1 <= M1Y1 or X1 - M1Y1 <= 0
X2 <= M2Y2 or X2 - M2Y2 <= 0
X3 <= M3Y3 or X3 - M3Y3 <= 0
If Xi > 0 these constraints force the associated Yi to equal 1.
If Xi = 0 these constraints allow Yi to equal 0 or 1, but the objective will cause Solver to choose 0.
Note that Mi imposes an upper bounds on Xi.
It helps to find reasonable values for the Mi.

31. Finding Reasonable Values for M1
Consider the resource constraints
2X1 + 3X2 + 6X3 <= 600 } machining
6X1 + 3X2 + 4X3 <= 300 } grinding
5X1 + 6X2 + 2X3 <= 400 } assembly
What is the maximum value X1 can assume?
Let X2 = X3 = 0
X1 = MIN(600/2, 300/6, 400/5)
= MIN(300, 50, 80)
= 50
Maximum values for X2 & X3 can be found similarly.

32. Summary of the Model MAX: 48X1 + 55X2 + 50X3 - 1000Y1 - 800Y2 - 900Y3
S.T.: 2X1 + 3X2 + 6X3 <= 600 } machining
6X1 + 3X2 + 4X3 <= 300 } grinding
5X1 + 6X2 + 2X3 <= 400 } assembly
X1 - 50Y1 <= 0
X2 - 67Y2 <= 0 linking constraints
X3 - 75Y3 <= 0
All Yi must be binary
Xi >= 0, i = 1, 2, 3

33. Potential Pitfall
Do not use IF( ) functions to model the relationship between the Xi and Yi.
Suppose cell A5 represents X1
Suppose cell A6 represents Y1
You’ll want to let A6 = IF(A5>0,1,0)
This will not work with Solver!
Treat the Yi just like any other variable.
Make them changing cells.
Use the linking constraints to enforce the proper relationship between the Xi and Yi.

34. Implementing the Model
See file Fig6-21.xls

35. End of Chapter 6